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Solutions Test paper 09 - Phase Test 2 - Differentiation
Section A
(1) Differentiate the following trigonometric functions with respect to x
. (a) sec3x (b) 4cot2x (c) cosec2x
(6 Marks)
(a) u=secx, du/dx=secxtanx, dy/du = 3u2, dy/dx = 3sec3x.tanx
(b) y =4cot2x, dy/dx = -8cosec22x
(c) u =cosecx, du/dx =-cosecx.cotx, dy/du =2u dy/dx =-2 cosec2x.cotx
(2) Differentiate the exponential and log functions with respect to x
(a) 2e0.4x (b)e(x3 +5) (c) (3e3x)2 (d) y = (ln4x)-1 (e) y = (x-3 - lnx-2) (8 Marks)
(a) (0.8)e0.4x (b)3x2e(x3 +5) (c) u = (3e3x) du/dx=9e3x, dy/du=2u, dy/dx= 18e6x
(d) u =ln4x, du/dx =1/x,dy/du = -1/u2, dy/dx=-1/x(ln4x)2 (e) (-3x-4+2/x)
(3) Use calculus to find the coordinates of the turning points of the following
functions and determine whether they are maximum or minimum.
(a) y = (2x-1)(3x+3) (b) y = (x2 +1)(x-1) (c) y =3x3 – 3x + 4
(9 marks)
(a) y = 6x2+3x- 3, dy/dx =12x+3 =0. x= -0.25, y = -3.375, d2y/dx2=12 Min
(b) y=( x3 - x2 +x -1) dy/dx =3x2 - 2x+1=0 = No SOLUTION
© dy/dx = 9x2 -3 = 0. x = +/ – (1/3)0.5, x=+0.577, y=2.85,x=-0.577, y=5.15,
d2y/dx2 = 18x, x =-(0.57) max, x = (0.57) min
(4) Use product rule to differentiate the following expressions wrt x
(a) y = x3sin2x (b) y = 2xe2x (c) y = exln7x
(9 Marks)
(a) u=x3, du/dx =3x2. v=sin2x, dv/dx = 2cos2x.dy/dx = 3x2sin2x+2x3cos2x
(b) u=2x, du/dx =2, v= e2x , dv/dx = 2e2x dy/dx =4x e2x + 2ex = 2ex(1 +2x)
© u=ex, du/dx=ex, v=ln7x, dv/dx = 1/x. dy/dx = (ex/x + ex ln7)
(5) Use quotient rule to differentiate the following expressions wrt x
(a) y =3x3 /2sin3x (b) y = 2x2.5/3e2x (c) y = e3x/ln5x
(9 Marks)
3
2
(a) u = 3x , du/dx = 9x , v=2sin3x, dv/dx =6cos3x,
dy = 18sin3x.x2 – 18x3cos3x = 4.5x2cosec3x(1 –xcot3x)
dx
4(sin3x)2
2.5
(b) u =2x , du/dx = 5x1.5, v = 3e2x,dv/dx=6e2x, dy/dx = 15e2xx1.5 - 12x2.5e2x
9e4x
1.5
2x
dy/dx = x (5 - 4x)/3e
© u=e3x, du/dx = 3e3x, v=ln5x, dv/dx =1/x,
dy/dx= e3x(3ln5x -1/x)
(ln5x)2
(6) Produce an expression for dy for each of the following :
dx
(a) 2x2 +y2 = 7 (b) x = 4y2 +2y (c) x = 5cos2Ө, y = 5sinӨ
(a) d(2x2) + dy2 = d7 , let u =y2 du=2y. du = d(y2) = du x dy
dx
dx dx
dy
dx dx
dy dx
ie 4x +2ydy = 0 ie dy = -2x
dx
dx
y
(b) 1 = 8ydy + 2dy ie dy = . 1 . . = . 1 .
dx
dx
dx (2 +8y)
2(1+4y)
(9 marks)
© dx = -10sin2Ө, dy = 5cosӨ dy = 5cosӨ
dӨ
dӨ
dx -10sin2Ө
= -0.25cosecӨ
(7) If z = 4x2 + xy + 2y2 determine dz and dz when x= 2.5 and y =1.2
dx
dy
(5 marks)
dz = 8x + y = 21.2
dz = x + 4y = 7.3
dx
dy
Section B
(8) y = 4x³ + px² + q has a turning point at (2,-1). Find the values of p and q
and hence the coordinates of the max/min turning points on the curve
.
(6 Marks)
Ans dy/dx =0 = 12x2 +2px, ie 48 + 4p =0. p = -12
-1 = 32 –48 +q. q = 15
dy/dx = 12x2 - 24x = 0 = x(x-2). x=0,y =15, x=2, y =-1
d2y/dx2 = 24x - 24 . x=0, max. x= 2, min.
(9) If current I = (V/R) , where V = 15 volts and R = 5 ohms, calculate the
change in current I when V is increased by 0.05 volts and R is decreased by
0.5 ohms
(4 Marks)
dI = dI dV + dI.dR = 1 (0.05) - (-V)(0.5) = 0.01 + 15(0.5) = 0.31A
dt dV dt dR dt
R
R2
5x5
(10)The formula for the volume V of a cylinder is V = πr2h where the radius r
is 0.12m and the height h is 0.15m
Determine the rate of change of volume when r decreases by 0.0015m/sec
and h increases by 0.001m/sec
(5 Marks)
2
dV = dV.dr + dV.dh = 2πrh(-0.0015) + πr (0.001) =
dt
dr dt dh dt
= (-1.696 x10-4 + 4.52x10-5 ) = -1.244 x 10-4 V/s
(11) Calculate the minimum area of metal plate required to make a length of
.
rectangular ventilation trunking of length 25m and enclosing a volume
of 6.25m3 Assume that the trunking is closed at both ends.
(6marks)
V = 6.25m3 = 25bh, h = (0.25/b)
A = 2bh + 50b + 50h = 2bh + 50b + 50(0.25/b) dA = 0 + 50 –12.5 . b=0.5=h
db
b2
Area = 0.5 + 25 + 25 =50.5m2
(12) A voltage Vs of 12v is applied across a series RC circuit where R = 10kΩ,
C = 10μF.
The charge, Q coulombs, on the capacitor is given by Q = CVc where Vc is
the voltage across the capacitor. Vc = Vs( 1-e-t/RC).
(a) Determine the charge Q after 0.12 second.
(b) If the current i = dQ = C dVc determine i after 0.2sec
.
dt
dt
(6marks)
(a) Q = 10-5. 20(1- e-1.2) Q = 1.398x 10-4C =139μC
(b) I = dQ/dt = Vs(e-t/RC) = 20(e-2) = 270.67μA
R
10000
(13) (a)The curve y =ax +b passes through the points (-1,-14) and (5,46).
x
(a) Calculate the values of a and b
(b) Find the stationary points, distinguishing between the maximum and
minimum point.
(6marks)
Ans -14 = -a - b
14 =
a+b
46 = 5a +b/5
230 = 25a + b a =9, b =5
y = 9x + 5/x, dy/dx =(9 – 5/x2 )=0, (x= 0.745,y =13.42) (x=-0.745, y=-13.42)
d2y/dx2 = (10/x3). (x= 0.745 min ) (x =-0.745 max)